# Homework Help: Electrostatics [Electric Field]

1. Nov 29, 2011

### format1998

1. The problem statement, all variables and given/known data

Point P is 5.00 cm from the origin and each charge is 20.0 cm from the origin.

Q1= -50.0 nC (20.0 cm from the origin on the -x axis)
Q2= +30.0 nC (20.0 cm from the origin on the +x axis)
Point P is 5.00 cm from the origin (in the +x axis)

2. Relevant equations

E = kQ/r2
ENET=E1+E2

3. The attempt at a solution

E1 = (8.99E9 Nm2/C2)((-50E-9 C)/(.25m)2) = -7192 N/C

E2 = (8.99E9 Nm2/C2)((+30E-9 C)/(.15m)2)= 11986.7 N/C

ENET=ENET=E1+E2=-7192 N/C + 11986.7 N/C = 4794.67 N/C, -x

According to the answer key, the answer is 19200 N/C, -x. What am I not understanding?

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2. Nov 29, 2011

### Staff: Mentor

To me it looks like the diagram has 20cm for the left charge, but your equations use 25cm. Am I misreading your diagram?

3. Nov 29, 2011

### format1998

I need to find the magnitude of the Electric Field at point P. The charge on the left (-50 nC) is 20 cm from the origin, point P is another 5 cm to the right of the origin. So, the -50 nC charge is 25 cm from point P.